Notice that if $a=\frac{2 \cdot 5^{m}+10}{3^{m}+1}$ and $b=\frac{9^{m}+1}{5^{m}+5}$ are both integers, then their product,
$$a b=\frac{\left(2 \cdot 5^{m}+10\right)\left(9^{m}+1\right)}{\left(3^{m}+1\right)\left(5^{m}+5\right)}$$
is also an integer.
Collecting now a factor of two in the numerator, this expression can be rewritten as
$$\frac{2\left(5^{m}+5\right)\left(9^{m}+1\right)}{\left(3^{m}+1\right)\left(5^{m}+5\right)}=\frac{2\left(9^{m}+1\right)}{3^{m}+1} .$$
Let us now see for which $m$ this expression is an integer. Noting that $9^{m}+1=\left(3^{m}+1\right)^{2}-2 \cdot 3^{m}$, we can further transform the product $a b$ into the form
$$a b=2 \cdot \frac{\left(9^{m}+1\right)}{3^{m}+1}=2 \cdot \frac{\left(3^{m}+1\right)^{2}-2 \cdot 3^{m}}{3^{m}+1}=2\left(3^{m}+1\right)-4 \frac{3^{m}}{3^{m}+1}$$
which is an integer if and only if $4 \frac{3^{m}}{3^{m}+1}$ is an integer.
We now prove that $2<4 \frac{3^{m}}{3^{m}+1}<4$.
Indeed, since $m \geq 1$, we have $3^{m} \geq 3$, and therefore $2 \cdot 3^{m}+2<2 \cdot 3^{m}+2 \cdot 3^{m}<4 \cdot\left(3^{m}+1\right)$; dividing these three terms by $3^{m}+1$ we then find $2<4 \frac{3^{m}}{3^{m}+1}<4$ as desired.
Therefore, if $4 \frac{3^{m}}{3^{m}+1}$ is an integer, it must necessarily be equal to 3, since it is strictly between 2 and 4.
Imposing this equality we obtain $4 \cdot 3^{m}=3 \cdot\left(3^{m}+1\right) \Rightarrow 3^{m}=3 \Rightarrow m=1$.
Therefore, if there exists an $m$ with the required property, it must be equal to 1. On the other hand, it is easy to verify that $\frac{2 \cdot 5^{1}+10}{3^{1}+1}=5$ and $\frac{9^{1}+1}{5^{1}+5}=1$ are both integers. We can therefore conclude that $m=1$ is indeed a solution and is the only one.